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Math Help - Integral convergence proof.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Integral convergence proof.

    Prove that the integral \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx is converges.


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    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove that the integral \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx is converges.


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    \left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\int^{t}_3\frac{\ln x}{x^2\sqrt{x^2-4}}dx

    Now  \ln x<x and  \frac{1}{\sqrt{x^2-4}}<\frac{1}{\sqrt{x}} \; \forall \; x\geq3

    Thus we get \left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\int^{t}_3 \frac{x}{x^2\cdot\sqrt{x}}dx=\int^{t}_3\frac{1}{x^  {3/2}}dx = \frac{2}{\sqrt{3}}-\frac{2}{\sqrt{t}}

    So finally  \lim_{t\to\infty}\left|\int^{t}_3\frac{\sin x\cdot \ln x}{x^2\sqrt{x^2-4}}dx\right|<\lim_{t\to\infty}\left(\frac{2}{\sqrt  {3}}-\frac{2}{\sqrt{t}}\right)=\frac{2}{\sqrt{3}}
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Prove that the integral \int^{\infty}_3\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx is converges.


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    Maybe...

    \int^{\infty}_3|\frac{sinx\cdot lnx}{x^2\sqrt{x^2-4}}dx|<\int^{\infty}_3\frac{1\cdot (x)}{x^2\sqrt{x^2-4}}dx<\int^{\infty}_3\frac{1}{x\sqrt{x}}dx<\int\fr  ac{1}{x^{1.5}}dx


    and so on...
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    MHF Contributor Also sprach Zarathustra's Avatar
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    I late!
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